Define a spherical shell: $S = \{ (x, y, z) \in \mathbb{R}^3 \big | 2 \leq x^2 + (y - 3)^2 + (z + 1)^2 \leq 4 \}$ What is the triple integral of $f(x, y, z)$ over $S$ in spherical coordinates? Assume that $x$, $y$, and $z$ are expressed in terms of $\rho$, $\theta$, and $\varphi$. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y + 3, z - 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y + 3, z - 1) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y - 3, z + 1) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y - 3, z + 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$
The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the concentric spheres with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. Here, the region $S$ ranges from a radius of $2$ to a radius of $4$. Therefore, we want $2 < \rho < 4$. $ \int_0^\pi \int_0^{2\pi} \int_2^4 \cdots \, d\rho \, d\theta \, d\varphi$ Although $S$ is a unit sphere, it's also shifted so that it's centered at $(0, 3, -1)$. How can we can keep the bounds for the sphere and integrate over $S$ ? We can add $3$ to $y$ everywhere to transform $S$ three units to the right. We write $f(x, y + 3, z)$ to account for this shift. This kind of a shift has no effect with change of variables, because its Jacobian is $1$. We can do the same for $z$. So our integral looks like this now: $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y + 3, z - 1) \cdots \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_2^4 f(x, y + 3, z - 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$